Solveeit Logo
Login

Question

Question: If \[x\sec \theta = 1 \pm y\tan \theta \] and \[{x^2}{\sec ^2}\theta = 5 + {y^2}{\tan ^2}\theta \] t...

If xsec⁡θ=1±ytan⁡θx\sec \theta = 1 \pm y\tan \theta and x2sec⁡2θ=5+y2tan⁡2θ{x^2}{\sec ^2}\theta = 5 + {y^2}{\tan ^2}\theta then the relationship between ′x′'x' and ′y′'y' is
A. x2+y2=9{x^2} + {y^2} = 9
B. x2y2+4(x2−y2)=0{x^2}{y^2} + 4({x^2} - {y^2}) = 0
C. x2y2−9y2=0{x^2}{y^2} - 9{y^2} = 0
D. x2y2+4x2−9y2=0{x^2}{y^2} + 4{x^2} - 9{y^2} = 0

Explanation

Solution

To solve this problem we need to find the value of the variables by using the two given equations. Substitute the values of one equation in another equation and use the formulas to simplify the equation to get the required answer.

Formula used: sec⁡2θ=(1+tan⁡2θ){\sec ^2}\theta = (1 + {\tan ^2}\theta )
1sec⁡θ=cos⁡θ\dfrac{1}{{\sec \theta }} = \cos \theta
(a−b)2=a2−2ab+b2{(a - b)^2} = {a^2} - 2ab + {b^2}

Complete step-by-step answer:
We first consider the equation xsec⁡θ=1±ytan⁡θx\sec \theta = 1 \pm y\tan \theta
It can be written as follows when taking the 1 to left hand side.
xsec⁡θ−1=±ytan⁡θx\sec \theta - 1 = \pm y\tan \theta
By squaring on both sides we get
(xsec⁡θ−1)2=(±ytan⁡θ)2{(x\sec \theta - 1)^2} = {( \pm y\tan \theta )^2}
Firstly, (xsec⁡θ−1)2{(x\sec \theta - 1)^2} is in the form(a−b)2=a2−2ab+b2{(a - b)^2} = {a^2} - 2ab + {b^2}. Here a=xsec⁡θa = x\sec \theta andb=1b = 1.
By expanding (xsec⁡θ−1)2{(x\sec \theta - 1)^2} we get x2sec⁡2θ−2xsec⁡θ+1→(A){x^2}{\sec ^2}\theta - 2x\sec \theta + 1 \to (A)
Secondly, (±ytan⁡θ)2{( \pm y\tan \theta )^2} can be taken as y2tan⁡2θ→(B){y^2}{\tan ^2}\theta \to ({\rm B})
This is because when we square ±\pmwe always end up with a positive term.
From the (A) and (B) we have,
⇒x2sec⁡2θ−2xsec⁡θ+1=y2tan⁡2θ\Rightarrow {x^2}{\sec ^2}\theta - 2x\sec \theta + 1 = {y^2}{\tan ^2}\theta
Since it is given that x2sec⁡2θ=5+y2tan⁡2θ⇒x2sec⁡2θ−5=y2tan⁡2θ{x^2}{\sec ^2}\theta = 5 + {y^2}{\tan ^2}\theta \Rightarrow {x^2}{\sec ^2}\theta - 5 = {y^2}{\tan ^2}\theta we can use this in the above equation and reduce as,
⇒x2sec⁡2θ−2xsec⁡θ+1=x2sec⁡2θ−5\Rightarrow {x^2}{\sec ^2}\theta - 2x\sec \theta + 1 = {x^2}{\sec ^2}\theta - 5
⇒x2sec⁡2θ−2xsec⁡θ+1−x2sec⁡2θ=−5\Rightarrow {x^2}{\sec ^2}\theta - 2x\sec \theta + 1 - {x^2}{\sec ^2}\theta = - 5
Now, x2sec⁡2θ{x^2}{\sec ^2}\theta term will get cancelled as they have opposite signs.
⇒−2xsec⁡θ+1=−5\Rightarrow - 2x\sec \theta + 1 = - 5
Taking 1 to right hand side and solving,
⇒−2xsec⁡θ=−5−1\Rightarrow - 2x\sec \theta = - 5 - 1
⇒−2xsec⁡θ=−6\Rightarrow - 2x\sec \theta = - 6
The minus signs on both sides can now be cancelled and we also divide 6 on the right-hand side by the 2 from the left-hand side.
⇒xsec⁡θ=62\Rightarrow x\sec \theta = \dfrac{6}{2}
Dividing 6 by 2 leaves a quotient of 3 on the right hand side
⇒xsec⁡θ=3\Rightarrow x\sec \theta = 3
Since we know that1sec⁡θ=cos⁡θ\dfrac{1}{{\sec \theta }} = \cos \theta
⇒x=3×1sec⁡θ\Rightarrow x = 3 \times \dfrac{1}{{\sec \theta }}
⇒x=3cos⁡θ\Rightarrow x = 3\cos \theta
Now we can substitute x=3cos⁡θx = 3\cos \theta in xsec⁡θ=1±ytan⁡θx\sec \theta = 1 \pm y\tan \theta to find tan⁡2θ{\tan ^2}\theta .
⇒3cos⁡θsec⁡θ=1±ytan⁡θ\Rightarrow 3\cos \theta \sec \theta = 1 \pm y\tan \theta
As we have already seen, 1sec⁡θ=cos⁡θ\dfrac{1}{{\sec \theta }} = \cos \theta the above equation can be written as
⇒3×1sec⁡θ×sec⁡θ=1±ytan⁡θ\Rightarrow 3 \times \dfrac{1}{{\sec \theta }} \times \sec \theta = 1 \pm y\tan \theta
Cancelling 1sec⁡θ×sec⁡θ\dfrac{1}{{\sec \theta }} \times \sec \theta we get 1 and multiply 1 with 3, we get 3.
⇒3=1±ytan⁡θ\Rightarrow 3 = 1 \pm y\tan \theta
Taking 1 to left had side we solve as,
⇒3−1=±ytan⁡θ\Rightarrow 3 - 1 = \pm y\tan \theta
⇒2=±ytan⁡θ\Rightarrow 2 = \pm y\tan \theta
On squaring 2=±ytan⁡θ2 = \pm y\tan \theta on both sides we get 4=y2tan⁡2θ4 = {y^2}{\tan ^2}\theta
⇒4=y2tan⁡2θ\Rightarrow 4 = {y^2}{\tan ^2}\theta
⇒4y2=tan⁡2θ→(C)\Rightarrow \dfrac{4}{{{y^2}}} = {\tan ^2}\theta \to (C)
Now consider x2sec⁡2θ=5+y2tan⁡2θ{x^2}{\sec ^2}\theta = 5 + {y^2}{\tan ^2}\theta which is given in the question.
⇒x2(1+tan⁡2θ)=5+y2tan⁡2θ\Rightarrow {x^2}(1 + {\tan ^2}\theta ) = 5 + {y^2}{\tan ^2}\theta
We know that sec⁡2θ=(1+tan⁡2θ){\sec ^2}\theta = (1 + {\tan ^2}\theta )
⇒x2+x2tan⁡2θ=5+y2tan⁡2θ\Rightarrow {x^2} + {x^2}{\tan ^2}\theta = 5 + {y^2}{\tan ^2}\theta
By bringing all the terms to left hand side, we get
⇒x2+x2tan⁡2θ−5−y2tan⁡2θ=0\Rightarrow {x^2} + {x^2}{\tan ^2}\theta - 5 - {y^2}{\tan ^2}\theta = 0
Since tan⁡2θ{\tan ^2}\theta is a common term for both x2{x^2} as well as y2{y^2} we can further reduce the equation as follows.
⇒x2+(x2−y2)tan⁡2θ−5=0\Rightarrow {x^2} + ({x^2} - {y^2}){\tan ^2}\theta - 5 = 0
as we already have 4y2=tan⁡2θ\dfrac{4}{{{y^2}}} = {\tan ^2}\theta from equation (C) we can use it in the above equation.
⇒x2+(x2−y2)4y2−5=0\Rightarrow {x^2} + ({x^2} - {y^2})\dfrac{4}{{{y^2}}} - 5 = 0
On multiplying 4y2\dfrac{4}{{{y^2}}} with (x2−y2)({x^2} - {y^2}) and simplifying we get,
⇒x2+(4x2y2−4y2y2)−5=0\Rightarrow {x^2} + \left( {\dfrac{{4{x^2}}}{{{y^2}}} - \dfrac{{4{y^2}}}{{{y^2}}}} \right) - 5 = 0
⇒x2+4x2y2−4−5=0\Rightarrow {x^2} + \dfrac{{4{x^2}}}{{{y^2}}} - 4 - 5 = 0
⇒x2+4x2y2−9=0\Rightarrow {x^2} + \dfrac{{4{x^2}}}{{{y^2}}} - 9 = 0
Taking y2{y^2} as LCM we can find that
⇒y2x2+4x2−9y2y2=0\Rightarrow \dfrac{{{y^2}{x^2} + 4{x^2} - 9{y^2}}}{{{y^2}}} = 0
⇒x2y2+4x2−9y2=0\Rightarrow {x^2}{y^2} + 4{x^2} - 9{y^2} = 0.
Hence the solution.
If xsec⁡θ=1±ytan⁡θx\sec \theta = 1 \pm y\tan \theta and x2sec⁡2θ=5+y2tan⁡2θ{x^2}{\sec ^2}\theta = 5 + {y^2}{\tan ^2}\theta then the relationship between′x′'x' and ′y′'y' is
x2y2+4x2−9y2=0{x^2}{y^2} + 4{x^2} - 9{y^2} = 0.

Option D is the correct answer.

Note: In the question, students make a mistake that has to change the term with ±\pm as its sign, when we have to square the sign, it will always result in a positive term.
Because (+)×(+)=+( + ) \times ( + ) = + and (−)×(−)=+( - ) \times ( - ) = +
Some of the formula used in the trigonometric functions,
1cosec⁡θ=sin⁡θ\dfrac{1}{{\operatorname{cosec} \theta }} = \sin \theta
1cotθ=tan⁡θ\dfrac{1}{{{\text{cot}}\theta }} = \tan \theta